Friday, January 15, 2010

Problem 418: Triangle, Incircle, Inradius, Equal Tangent circles, Radius

Proposed Problem
Click the figure below to see the complete problem 418 about Triangle, Incircle, Inradius, Equal Tangent circles, Radius.

Problem 418: Triangle, Incircle, Inradius, Equal Tangent circles, Radius.
See also:
Complete Problem 418
Level: High School, SAT Prep, College geometry

3 comments:

  1. join A to D & to O, C to E & to O ( D & O on bisector )
    join D to E, meet r at P
    r meet AC at T

    mark FP = z, AT = y

    ▲DPO ~ ▲ATO, => (x+z)/y = (r-x)/r (1)

    ▲PEO ~ ▲TCO, => (x-z)/(b-y) = (r-x)/r (2)
    from (1) & (2)=>

    (x+z)/y = (x-z)/(b-y)
    xb-xy+zb-zy = xy-zy
    xb+zb = 2xy

    x+z = 2xy/b (3)

    substitute (3) at (1)

    (2xy/b)/y = (r-x)/r
    2x/b = (r-x)/r
    2xr = br-bx
    (2r+b)x = br

    x = br/(2r+b)

    ReplyDelete
  2. I the incenter
    the two circles touch AC at G and H
    AG+HC=b-2x
    [AIC]=[ADG]+[CEH]+[GDEH]+[DIE]
    br=x(b-2x)+4x²+2x(r-x)
    br=x(b+2r)
    x=br/(b+2r)
    .-.

    ReplyDelete
  3. Let L, M, N be the feet of the Perpendiculars from D, E, I upon AC respectively.
    Denote AL = y, CM = z.
    Note AN = s - a, CN = s - c
    DL // IN implies x:r = y:(s-a)
    EM // IN implies x:r = z:(s-c)
    From b = y + 2x + z, we have
    rb = ry + rz + 2rx = x(s-a)+ x(s-c) + 2rx
    => rb = x(b + 2r), x = br/(b +2r)

    ReplyDelete